Q.The coeff of x^3 y^4 z^5 in the expression of (xy+yz+xz)^6 is..?

1.70

2.60

solve for a,b,c

a=1

b=3

c=2

so v have

now consider

v need the term

so v use the term

v need which will have coeff = 5C3 from tht 't' equation

so coeff=6C1*5C3=60

here's a better method

general term in the expansion is

=

so r+t = 3

r+s = 4

s+t = 5

solving we get s = 3 , t = 2 , r = 1

so coeffcient is  

see dude...no pt in doin tht way.. dun u think it will be better to prove it...no one is posin the q for some1 to apply some std formula...

the std formula ne1 know is (x+y)^n...so i think it wud help if u proved wid tht.

hey dude i'm not using any standard formula

just as u know that general term of the expansion of (x+y)^n is n!/r!s! x^r y^s  where r + s = n

so extending this to three terms general term for (x+y+z)^n is n!/r!s!t! x^r y^s z^t where r + s + t = n ,

this is very basic my friend!!!!!

if im not wrong, thats a multinomial coefficient :D

unless u prove it its not basic :)

gnly v deal only wid binomial so u shud prefer to use tht while explainn

no the multinomial coefficient is quite basic :D

if ppl found tht basic they would not have posted the question

then the forum shdnt exist da :D :D But thats a different issue altogether :D Im saying that wrt an iit student it shd be quite basic and u knw it urself!

thts well known formula i accept...but then this question becomes direct...but its posted cuz the person not aware of formula...so deriving the answer will be the best thing to do :)

Wouldn't teaching the person the basic formula help too? :huh: Bcos the method will be shorter (i dont say ur method is bad ofc!)

no pt knowing formulae widout logic behind it...atleast thts the secret behind physics...derive everything

failry simple..

correctly done by priyesh..

The general term in the expansion of ({xy+yz}+xz)^6 is ..

..this is the (r+1)th term

now for the expansion of (xy+yz)^6-r..the gt is